Two particles bounce off each other and continue moving in different directions.

Question

Two particles bounce off each other and continue moving in different

directions. The first particle's direction is defined by ~v = 6^i-^j+^k and

the second particle's direction is defined by ~w = 2^i+ 2^j- ^k. Use the

vector product to find the angle between the paths of the two particles.

Show your working. Round the answer to 3 signicant gures.

Answer 1

Hello Ramon;

Given vectors: ~v = 6^i - ^j + ^k and ~w = 2^i + 2^j - ^k.

The magnitudes: |~v| = √38 and |~w| = 3.

The cross product: ~v x ~w = 12^i + 12^j - 1^k.

The magnitude of the cross product: |~v x ~w| = 17.

Sin(ø) = |~v x ~w| / (|~v| * |~w|) ≈ 0.592.

ø ≈ sin^-1(0.592) ≈ 36.3^o

I hope this short outline helps.

Answer 2

sinθ = IIv x wII/(IIvII·IIwII); v x w = det [(i j k) (6 -1 1) (2 2 -1)] = - i + 8j + 14k;

v • w = 12 - 2 - 2 = 8 > 0. So, θ is acute; sinθ = (1 + 64 + 196)^1/2/[(36 + 1 + 1)(4 + 4 + 1)]^1/2 = 0.8735891;

θ = 60.88°